0 Kelvin body moving As many books say: 

Temperature is (proportional, almost, etc...) average kinetic energy of particles.

My question is this.

"Suppose there is a body somewhere in empty space which moves at velocity v thus possessing KE not equal to 0 Joules, but has temperature equal to 0K. Can this be possible in a theoretical sense?"

In my view it is possible since I think T is a measure of KE only for non-coherently moving particles. And if all particles of the body move coherently in one direction KE will not be equal to 0 Joules, but T will be 0 Kelvin.
P.S. Since in the case of classical Thermodynamics, the engine (e.g. Carnot engine) does not move as a whole; it does not possess any "coherent KE", only "random KE" and thus the heuristic "temperature is average KE" works just fine. But this "rule of thumb" breaks down if the body moves as a whole. Am I right? 
 A: I think you are right. A perhaps more precise relation between temperature and velocity is the Maxwell–Boltzmann distribution:
\begin{equation*}
P(\textbf{v}) = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \text{exp} \left[-\frac{m ( \textbf{v} - \textbf{v}_0)^2}{2 k_B T} \right].
\end{equation*}
where you see that the mean velocity $\textbf{v}_0$ and the temperature are independent. Only the variance of the velocity is related to temperature.
A: It's simple. You can think of temperature as being the standard deviation of KE among all components (atoms) of a mass. This is significant because KE is a relative quantity, but temperature is absolute, and this relationship makes that possible. If all atoms are moving uniformly in the same direction, then the temperature would be 0.
A: I think your view is correct, and you can think about the following real word example.  In labs here on earth, we can use laser cooling techniques to cool atoms to $\mu$K scales in the lab frame.  But the lab is on earth, and the earth is moving very fast around the sun, and the sun is moving very fast around the galactic center and so on.  We don't take into account this additional kinetic energy when considering the temperature.
If your object is moving at a constant velocity, you can always put yourself in a frame where the mean velocity of that object is zero, and what you are left with is the distribution of velocities that will define the temperature of the object.
A: You are correct.  The temperature is usually defined in terms of random motion of the atoms.  This announcement from NIST talks of an atomic beam cooled to $30 \mu K$, so there is very little random energy.  It doesn't give the speed of the beam, but that could be very high if the atoms are all moving in the same direction.
A: One way to define temperature is via statistical physics. It is a quantity used to describe an equilibrium distribution of possible states via Boltzmann factor. Given a total energy of a classical system 
$$E(x_1,\ldots,x_N, p_1, \ldots p_N)$$,
where $x_1 \ldots x_N$ are the position coordinates, and $p_1 \ldots p_N$ are the momentum coordinates. The probability of the system being in a particular state in now given proportional to
$$P(x_1,\ldots,x_N, p_1, \ldots p_N) = Z^{-1} e^{-E(x_1,\ldots,x_N, p_1, \ldots p_N) / (k_B T)}$$
In classical system of particles, it almost always is the case, that the total energy is a sum of kinetic energy and velocity independent potential energy.
$$E(x_1,\ldots,x_N, p_1, \ldots p_N) = V(x_1,\ldots,x_N) + \sum_i^{N} \frac{p_i^2}{2m} $$
We could play with the idea of dividing the total kinetic energy of the system to three components, translation kinetic energy, rotation kinetic energy and vibrational kinetic energy with help of suitable coordinate system
$$E_{tot} = E_{trans} + E_{rot} + E_{vib}$$,
where the translation kinetic energy is given by
$$E_{trans} = \frac{P^2}{2M}$$,
where $P=\sum_i p_i$ and $M=\sum_i m_i$. Now, we can speculate, that the probability for the system center of mass having a certain velocity V is roughly proportional to
$$P(V) \approx e^{-\frac{1}{2}MV^2 / (k_BT)}.$$
Given that, the total kinetic energy is huge compared to $k_BT$ (which is just fractions of an eV in room temperaure), we can conclude that if we find a system with large V, this is a non-equilibrium system and temperature is defined only on equilibrium systems. Therefore, the definition of temperature is ill-defined here.
Ok, this was just purely formal. In practice, one does not consern one-self with such formalities, but just moves to center-of-momentum coordinates or something similar. For example, it is standard practice in molecular dynamics simulations to first initialize the velocities of atoms according to Maxwell-Boltzmann distribution, but then to remove total momentum and total angular momentum to prevent excess drift and rotation. The number of particles is usually so large, that one does not even consider the fact that there are few degrees of freedom less, than one would get just by based on particle number. An exception is few particle systems, where one needs to switch to reduced mass coordinates.
